Adding sequence of square roots

Question

How to add sequence of square roots from square root 2 till square root 99 and how to add the sequence of their reciprocal here is the original problem

Answer 1

The original sums you asked for have no nice closed form and so a calculator is going to be the only feasible way to get a numerical result.

The sum you link to in the image is a totally different one and will have a nice result.

Notice that $\frac{1}{\sqrt{n}+\sqrt{n+1}} = \frac{1}{\sqrt{n}+\sqrt{n+1}}\cdot \frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}} = \frac{\sqrt{n}-\sqrt{n+1}}{n-(n+1)}=(\sqrt{n+1} - \sqrt{n})$

You have as a result:

$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots+\frac{1}{\sqrt{99}+\sqrt{100}}$

$=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\dots+(\sqrt{100}-\sqrt{99})$

$=\sqrt{100}-\sqrt{1} = 9$

Answer 2

By the way if any one is looking for the sum of sequence of square roots here is the answer

http://ramanujan.sirinudi.org/Volumes/published/ram09.pdf